Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i) \({{1} \over {4}} , -1\)
\(\Rightarrow \)Sum of the zeroes = \({{1} \over {4}}\) = \({{-b}\over {a}}\)
\(\Rightarrow \) Product of the zeroes = -1 = \({{c}\over {a}}\)
Polynomial can be formed by : \(x^2\) - (sum of zeroes)x + (product of zeroes)
\(\Rightarrow \)\(x^2 - {{1} \over {4}}x + (-1)\) => \(4x^2 - x - 4\)

(ii) \( \sqrt{2} , {{1}\over{3}}\)
\(\Rightarrow \)Sum of the zeroes = \( \sqrt{2}\) = \({{-b}\over {a}}\)
\(\Rightarrow \) Product of the zeroes = \({{1} \over {3}}\) = \({{c}\over {a}}\)
Polynomial can be formed by : \(x^2\) - (sum of zeroes)x + (product of zeroes)
\(\Rightarrow \)\(x^2 - \sqrt{2} x + {{1} \over {3}}\) \(\Rightarrow \) \(3x^2 - 3\sqrt{2}x + 1\)

(iii) \(0 , \sqrt{5}\)
\(\Rightarrow \)Sum of the zeroes = 0 = \({{-b}\over {a}}\)
\(\Rightarrow \) Product of the zeroes = \(\sqrt{5}\) = \({{c}\over {a}}\)
Polynomial can be formed by : \(x^2\) - (sum of zeroes)x + (product of zeroes)
\(\Rightarrow \)\(x^2 - 0x + ( \sqrt{5})\) => \(x^2 + \sqrt{5}\)

(iv) \(1 , 1\)
\(\Rightarrow \)Sum of the zeroes = 1 = \({{-b}\over {a}}\)
\(\Rightarrow \) Product of the zeroes = 1 = \({{c}\over {a}}\)
Polynomial can be formed by : \(x^2\) - (sum of zeroes)x + (product of zeroes)
\(\Rightarrow \) \(x^2 - 1x + 1\)

(v) \({{-1} \over {4}} , {{1} \over {4}}\)
\(\Rightarrow \) Sum of the zeroes = \({{-1} \over {4}}\) = \({{-b}\over {a}}\)
\(\Rightarrow \) Product of the zeroes = \({{1} \over {4}}\) = \({{c}\over {a}}\)
Polynomial can be formed by : \(x^2\) - (sum of zeroes)x + (product of zeroes)
\(\Rightarrow \) \(x^2 - {{-1} \over {4}}x + {{1} \over {4}}\) \(\Rightarrow \) \(4x^2 + x + 1\)

(vi) \(4 , 1\)
\(\Rightarrow \) Sum of the zeroes = 4 = \({{-b}\over {a}}\)
\(\Rightarrow \) Product of the zeroes = 1 = \({{c}\over {a}}\)
Polynomial can be formed by : \(x^2\) - (sum of zeroes)x + (product of zeroes)
\(\Rightarrow \) \(x^2 - 4x + 1\)